22,209
社区成员
发帖
与我相关
我的任务
分享
select cast(dt as date) - interval extract(day from dt)-1 day from(select now() as dt) a;
--测试数据
if not object_id(N'Tempdb..#T') is null
drop table #T
Go
Create table #T([user_id] int,[money] int,[data] Date)
Insert #T
select 12333,100,'2018/7/1' union all
select 12333,101,'2018/7/6' union all
select 12333,102,'2018/7/9' union all
select 12333,100,'2018/7/15' union all
select 12333,101,'2018/7/28'
Go
--测试数据结束
SELECT
MONTH(data) AS 月份,
SUM(money) AS summoney
FROM
#T
WHERE
DATEDIFF(MONTH, data, '2018-08-01') = 1
GROUP BY
MONTH(data);